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2 years ago

NEED AN ANSWER ASAP!!! WILL REWARD HANDSOMELY!!!

Mathematics

2 answers:

Mrrafil [7]2 years ago

5 0

I think the answer will be 5?

Ahat [919]2 years ago

4 0

Answer:

See attached diagram

We have

GT = 9.5 ft
BT = 11 ft

Let's find angle x and compare with the safety requirement

GT/BT = sin x
9.5/11 = sin x
x = arcsin (9.5/11)
x = 59.7° < 75°

The angle we got is less than 75° so it is safe to use

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How to find the volume of a quarter-sphere?

Ganezh [65]

As the word suggests, a quarter-sphere is a quarter of a sphere, and thus the volume of a quarter-sphere is a quarter of the volume of the sphere.

The volume of a sphere is computed as

$\dfrac{4}{3} \pi r^3$

So, the volume of a quarter-sphere is

$\dfrac{1}{4}\cdot \dfrac{4}{3} \pi r^3 = \dfrac{1}{3} \pi r^3$

7 0

2 years ago

Help please!! Thank you!! :)

Naddik [55]

Answer:

A. m^2+7m+10=0

Step-by-step explanation:

This is a problem in pattern matching, and in substituting a variable for a pattern.

(x^2+3)^2 +7x^2 +21 = -10 . . . . . . given

(x^2 +3)^2 +7(x^2 +3) = -10 . . . . . factor the last two terms

m^2 +7m = -10 . . . . . . . . . . . . subsitute m for x^2 +3

m^2 +7m +10 = 0 . . . . . . . . add 10 to both sides; matches A

5 0

1 year ago

50 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Sliva [168]

Answer:

Step-by-step explanation:

18^2 + 80^2 = 82

324 + 160 = c^2

4 0

2 years ago

Read 2 more answers

Graph g(x) = x2 +2 Compare the graph to the graph of f(x) = x2.

Olegator [25]

Step-by-step explanation: Remember that the parent graph f(x) = x², is a parabola that opens up and passes through the origin.

Notice that g(x) = x² + 2 has 2 added to the parent term.

This means that g(x) = x² + 2 is the graph of f(x) = x² translated two units up.

5 0

2 years ago

A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s

mojhsa [17]

Answer:

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, $\theta=25$ °

Initial speed is, $u=15\ m/s$

Time of flight is, $t=3.0\ s$

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

$u_y=u\sin\theta\\\\u_y=15\sin(25)=6.34\ m/s$

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = $-h$ (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is, $g=-9.8\ m/s^2$

Now, using the following equation of motion;

$-h=u_yt+\frac{1}{2}gt^2\\-h=6.34(3)-\frac{9.8}{2}(3)^2\\\\-h=19.02-44.1\\\\-h=-25.08\\\\h=25.08\ m$

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So, $v_y=0\ m/s$

Now, using the following equation of motion, we have:

$v_y^2=u_y^2+2gH\\\\0=(6.34)^2-2\times 9.8\times H\\\\19.6H=40.2\\\\H=\frac{40.2}{19.6}=2.05\ m$

Therefore, the maximum height of the brick is 2.05 m.

5 0

3 years ago